Uniform and Exponential Distribution of Random Variables

Date: 01/16/2006 at 16:49:00
From: troy
Subject: Relationship between uniform distrib and exponential distrib
I've seen statements claiming if you take the natural log of a
uniformly distributed random variable, it becomes a exponentially
distributed random variable. I know it's a true statement, but I
wonder if you would provide a proof?

Date: 01/17/2006 at 12:26:42
From: troy
Subject: Thank you (Relationship between uniform distrib and
exponential distrib)
Thank you, Doctor George, for the succinct proof. I'd like to
continue the discussion below. Although it's not a complex
derivation, it's definitely non-trivial. I wondered why most articles
skipped the proof and only made the statement. Could I ask you to
point me to a reference where I could find similar derivations?
Also on the last equation, f(y) = f[e(-y)]e(-y) = e(-y),
Y X
it implies that f[e(-y)] = 1.
X
What is the reason for this? Is it because by definition that Y =
-ln(X), therefore x = e(-y), so the probability of x = e(-y) is always 1?
Thank you again for the beautiful derivation.
Troy

Date: 01/17/2006 at 13:58:33
From: Doctor George
Subject: Re: Thank you (Relationship between uniform distrib and
exponential distrib)
Hi Troy,
You are on the right track. Since X is uniform,
f(x) = 1
X
now substituting x = e(-y) we get
f[e(-y)] = 1
X
As for a reference, most any college level book on Mathematical
Statistics will contain examples similar to this. Look for a section
on transformation of variables. Sometimes a book will skip a short
proof like this and have the reader work it out as a problem.
- Doctor George, The Math Forum
http://mathforum.org/dr.math/